In an acute triangle ABC, if the coordinates of orthocenter H are (4, b), of centroid G are (b,2b - 8), and of circumcenter C are (-4,8), then b can not be

As F1 (orthocenter), G (centroid), and C (circumcenter) are collinear, we have

$\begin{array}{l}\left|\begin{array}{ccc}4& \mathrm{b}& 1\\ \mathrm{b}& 2\mathrm{b}-8& 1\\ -4& 8& 1\end{array}\right|=0\\ \mathrm{or} \left|\begin{array}{ccc}4& \mathrm{b}& 1\\ \mathrm{b}-4& \mathrm{b}-8& 0\\ -(\mathrm{b}+4)& 16-2\mathrm{b}& 0\end{array}\right|=0\\ \mathrm{or} (\mathrm{b}-4)(16-2\mathrm{b})+(\mathrm{b}+4)(\mathrm{b}-8)=0\\ \mathrm{or} 2(\mathrm{b}-4)(8-\mathrm{b})+(\mathrm{b}+4)(\mathrm{b}-8)=0\\ \mathrm{or} (8-\mathrm{b})\left[\right(2\mathrm{b}-8)-(\mathrm{b}+4\left)\right]=0\\ \mathrm{or} (8-\mathrm{b})(\mathrm{b}-12)=0\end{array}$

Hence, b = 8 or 12, which is wrong because collinearity does not
explain centroid, orthocenter, and circumcenter.
Now, H., G, and C are collinear and $\frac{\mathrm{HG}}{\mathrm{GC}}=2$. Therefore,

$\begin{array}{r}\frac{-8+4}{3}=\mathrm{b}\text{ or }\mathrm{b}=\frac{-4}{3}\\ \mathrm{and} \frac{16+\mathrm{b}}{3}=2\mathrm{b}-8\text{ or }\mathrm{b}=8\end{array}$

But no common value of b is possible.